(by Hyp.), and ok is greater than ox (I. 17.), the rectangle ov will also be greater than the polygon Inrt.

But the polygon differs from the circle, or from the rectangle oz, by lefs than the magnitude wz (by Conft.), and ov differs from oz by wz; confequently the polygon is greater than the rectangle ov.

It is, therefore, both greater and lefs at the fame time, which is abfurd; whence the circle kmps is not greater than the rectangle ov.

Again, let it be less than ov, by the rectangle wy; and let BDFH be a polygon circumfcribed about the circle, that fhall differ from it by lefs than the magnitude wy (VIII. 4.)

Then fince the triangle BOA is equal to half a rectangle under the base BA and the perpendicular ok (I. 32.), the whole polygon will be equal to half a rectangle under its perimeter and perpendicular ok.

And because ow is lefs than half the perimeter of any polygon that can be circumscribed about the circle (by Hyp.), and ok is common, the rectangle ov will also be lefs than the polygon BDFH.

But the polygon differs from the circle, or from oy, by lefs than the magnitude wy (by Hyp.), and ov differs from oy by wy; confequently the rectangle ov will be greater than the polygon BDFH, which is abfurd.

Since, therefore, the rectangle ov is neither greater nor less than the circle kmps, it must be equal to it, as was to be fhewn.

PRO P. VII. THEOREM.

The circumferences of circles are in proportion to each other as their diameters.

S

G

Let ABCD, EFGH be any two circles, whofe diameters are BD, FH; then will the circumference ABCD be to the circumference EFGH as the diameter BD is to the diameter FH.

For let OM, SP be two right lines equal to the femicircumferences DAB, HEF, and on the radii OA, SE make the fquares OK, SL (II. 1.), and complete the rectangles

ON, SR:

Then fince the rectangles ON, SR are equal to the circles ABCD, EFGH (VIII. 6.), and the circles are to each other as the fquares of their radii (VIII. 5. Cor.) the rectangle on will alfo be to the fquare ok as the rectangle SR is to the fquare SL (V. 9.)

But on is to OK as Oм to OD (VI. 1.), and SR to SL as SP to sH (VI. 1.); therefore, by equality, om will be to OD as SP is to SH (V. 11.)

And because any equimultiples of four proportional quantities, are also proportional (V.13.), twice Om will be to twice OD as twice sp is to twice SH; or, by alternation, twice Oм is to twice SP as twice OD is to twice.

But twice oм and twice SP are equal to the circumferences ABCD, EFGH (by Conft.); and twice OD and twice SH are equal to the diameters BD, FH; whence the circumference ABCD is to the circumference EFGH as the diameter BD is to the diameter FH.

PROP. VIII. THEOREM.

Q. E. D.

If a prifm be cut by a plane parallel to its base, the section will be equal and like the bafe.

Let AG be a prifm, and KLMN a plane parallel to the base ABCD ; then will KLMN be equal and like ABCD. For join the points NL, and DB:

Then fince KM, AC are parallel planes (by Hyp.), and the plane AN cuts them, the section KN will be parallel to the fection AD (VII. 12.)

And fince AK is also parallel to DN (VIII. Def. 3.), the figure AN is a parallelogram; and consequently KN is equal to AD (I. 30.)

In like manner it may alfo be fhewn, that KL is equal to AP, LM to BC, and MN to CD.

And fince KN, KL in the plane KM, are parallel to AD, AB in the plane AC, the angle NKL will be equal to the angle DAB (VII. 7.)

The two fides KN, KL of the triangle KLN, being, therefore, equal to the two fides AD, AB of the triangle ABD, and the angle NKL to the angle DAB, the triangle KLN will be equal and like the triangle ABD (I. 4.)

And in the fame manner it may be fhewn, that the triangle LMN is equal and like to the triangle BCD.

But the triangles KLN, LMN are, together, equal to the section KLMN; and the triangles ABD, BCD to the fection ABCD; whence the fection KLMN is equal and like to the fection ABCD.

Q. E. D.

PROP. IX. THEORE M.

Prisms of equal bafes and altitudes are equal to each other.

Let AM, ES be any two prifms, ftanding upon the equal bafes ABCD, EFGH, and having equal altitudes; then will AM be equal to ES.

For parallel to the bafes, and at equal diftances from them, draw the planes mp and vw.

Then, by the laft propofition, the section mnps will be equal to the bafe ABCD, and the section vour to the base

EFGH.

1

But the base ABCD is equal to the bafe EFGH by hypothefis; whence the fection mnps is, alfo, equal to the fection vowr.

And in the fame manner it may be fhewn, that any other fections, at equal diftances from the bafes, are equal to each other.

Since therefore every fection in the prifm AM is equal to its corresponding fection in the prifm Es, the prifms themfelves, which are compofed of thofe fections, muft alfo be equal. Q. E. D. COR. Every prifm is equal to a rectangular parallelepipedon of an equal base and altitude.

PROP. X. THEOREM.

Rectangular parallelepipedons, of equal altitudes, are to each other as their bafes.

Let AC, MP be two rectangular parallelepipedons, having the equal altitudes ED, QR; then will AC be to MP as the base BE is to the base NQ.

For in AB, produced, take any number of right lines AF, FL each equal to AB; and in MN, produced, take any number of right lines NT, TX each equal to MN.

Complete the parallelograms FE, FK, MV, TZ, and make the upright folids AG, FH, NW, TY of equal altitudes with AC or MP.